Two-Pair probability

The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!

Re: Two-Pair probability

The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!

Re: Two-Pair probability

I'm not finished with the problem. I had typed out a response to you, but after submission read a more detailed response from Soroban, so I quickly edited away what I wrote. I haven't had a chance to look over his response yet as I just got home, but I will!

Re: Two-Pair probability

Ahh I see my mistake now, Soroban. So the general formula that I was using gives ordered partitions. Is there any such a formula for unordered partitions? It seems simple but I don't want to make a similar mistake on a similar problem later. Thank you for the help!